Javascript 使用HTML登录Mysqli php表单_Javascript_Php_Mysql_Mysqli

Javascript 使用HTML登录Mysqli php表单

javascript php mysql

Javascript 使用HTML登录Mysqli php表单,javascript,php,mysql,mysqli,Javascript,Php,Mysql,Mysqli,我有一个HTML登录表单和一个php代码，可以启动数据库连接并通过简单的HTML登录。我想做的是在我拥有的特定设计上实现php代码。我在将php代码与HTML表单所需的设计连接时遇到问题登录html表单： <li id="login"> <a id="login-trigger" href="#"> Log in <span>▼</span> </a> <div id="login-content">

我有一个HTML登录表单和一个php代码，可以启动数据库连接并通过简单的HTML登录。我想做的是在我拥有的特定设计上实现php代码。我在将php代码与HTML表单所需的设计连接时遇到问题

登录html表单：

<li id="login">
  <a id="login-trigger" href="#">
    Log in <span>▼</span>
  </a>
  <div id="login-content">
    <form>
      <fieldset id="inputs">
        <input id="username" type="email" name="Email" placeholder="Your email address" required>   
        <input id="password" type="password" name="Password" placeholder="Password" required>
      </fieldset>
      <fieldset id="actions">
        <input type="submit" id="submit" value="Log in">
        <label><input type="checkbox" checked="checked"> Keep me signed in</label>
      </fieldset>
    </form>
  </div>                     
</li>
<li id="signup">
  <a href="">Sign up FREE</a>
</li>

PHP MySQL代码包括一个HTML登录表单：

<html>
<head>

</head>
<body>

<?php
if (!isset($_POST['submit'])){
?>
<!-- The HTML login form -->
    <form action="<?=$_SERVER['PHP_SELF']?>" method="post">
        Username: <input type="text" name="username" /><br />
        Password: <input type="password" name="password" /><br />

        <input type="submit" name="submit" value="Login" />
    </form>
<?php
} else {
    require_once("db_const.php");
    $mysqli = new mysqli(DB_HOST, DB_USER, DB_PASS, DB_NAME);
    # check connection
    if ($mysqli->connect_errno) {
        echo "<p>MySQL error no {$mysqli->connect_errno} : {$mysqli->connect_error}</p>";
        exit();
    }

    $username = $_POST['username'];
    $password = $_POST['password'];

    $sql = "SELECT * from users WHERE username LIKE '{$username}' AND password LIKE '{$password}' LIMIT 1";
    $result = $mysqli->query($sql);
    if (!$result->num_rows == 1) {
        echo "<p>Invalid username/password combination</p>";
    } else {
        echo "<p>Logged in successfully</p>";
        // do stuffs
    }
}
?>      
</body>
</html>

在其他地方试试这个

! 解析错误：语法错误，第37yo行出现意外“}”，您必须在else{-this code-}中编写此代码！致命错误：在第19行调用未定义的函数mysqli调用堆栈时间内存函数位置1 0.0008 245368{main}..\2.php:0如果您希望在成功登录后执行此操作，则获得：php警告：无法修改标头信息-标头已由从开始的输出发送。使用上面的代码如何解决这个问题？

    require_once("db_const.php");
    $mysqli = mysqli_connect(DB_HOST, DB_USER, DB_PASS, DB_NAME);
    mysqli_select_db($mysqli,DB_NAME);
    # check connection
    if ($mysqli->connect_errno) {
        echo "<p>MySQL error no ".$mysqli->connect_errno." : ".$mysqli->connect_error."</p>";
        exit();
    }

    $username = $_POST['username'];
    $password = $_POST['password'];

    $sql = "SELECT * from users WHERE username = '".$username."' AND password = '".$password."' LIMIT 1";
    $result = $mysqli->query($sql);
    if (!$result->num_rows == 1) {
        echo "<p>Invalid username/password combination</p>";
    } else {
        echo "<p>Logged in successfully</p>";
        // do stuffs
    }